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given: △abc, bk=10, ac=30, m∠nmo=90°, mn = mo,
bk⊥ac, no∥ac, m∈ac

find: no

Step-by-step explanation:

I have a slightly different way to solve this.

Let P be the mid-point of NO.

As MN = MO, MNO is an isosceles triangle,

line MP is ⊥ to NO

As ∠NMO=90°, triangle MNO is similar to triangles PNM and POM.

MP = NP = PO

MP = NO/2

Let Q be the interaction of BK and NO

BQ = BK - QK

As QK = MP

BQ = BK - MP

= 10 - NO/2

Triangle BNO and BAC are similar

BQ/NO = BK/AC

Substituting BQ by 10 - NO/2

(10 - NO/2) / NO = 10 /30

NO = 3*(10 - NO/2)

NO = 30 - 3*NO/2

5*NO/2 = 30

NO = 12

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i

(-5,-28)

i graphed to on desmos.